queries query

alternative

this is my code

<form method="POST" action="<?$_POST['categoryID']?>">   <!--after the ID posted to abc.php, example it will query abc.php?categoriID=1  -->
<select> 
<?
// gonna connect to the database 1st. database = ddm
$conn = mysql_connect("localhost", "username", "password"); 
$db = mysql_select_db("ddm"); 


// running sql query  table = category, attributes = categoryID, cate
$list = mysql_query("SELECT * FROM category");  

while ($row = @mysql_fetch_assoc($list))
{ //option value everything is here, query as much as data from database
echo "<option value=\"$row[categoryID]\">$row[cate]</option>\n<br>";
}
if ($_POST['categoryID'] !=" ") 
{
	$conn = mysql_connect("localhost", "username", "password"); 
	$db = mysql_select_db("ddm"); 
	$list2 = mysql_query("SELECT * FROM category2 where categoryID=$categoryID");
	while ($row2 = @mysql_fetch_assoc($list2)) 
	{
	echo "<option value=\"$row2[categoryID]\">$row[cate2]</option>\n<br>"; 
	} 
}
?>
</select>
<input type="submit" value="query" name="B1"> <!-- submit button--> 
</form>

--------------------

i still dun understand why i cant query the queries.

thanx in advance

lanjoky


$conn = mysql_connect("localhost", "username", "password");  

$db = mysql_select_db("ddm");  


// running sql query  table = category, attributes = categoryID, cate 
$list = mysql_query("SELECT * FROM category", $conn); //change here   

while ($row = @mysql_fetch_assoc($list)) 
{ //option value everything is here, query as much as data from database 
echo "<option value=\"$row[categoryID]\">$row[cate]</option>\n<br>"; 
} 
if ($_POST['categoryID'] !=" ")  

{ 
    $conn = mysql_connect("localhost", "username", "password");// delete this line  

    $db = mysql_select_db("ddm");  // delete this line
    $list2 = mysql_query("SELECT * FROM category2 where categoryID=$categoryID", $conn);//change 
    while ($row2 = @mysql_fetch_assoc($list2))  

    { 
    echo "<option value=\"$row2[categoryID]\">$row[cate2]</option>\n<br>";  

    }  

}

Try that and see if it works ... I made comments near the lines to delete also put comment next to lines I changed

alternative

hm..
i am sorry..
this din shown any results.. i tested the code already

lanjoky

Where you have <form method="POST" action="<?$_POST['categoryID']?>">

Change it to:
<form method="POST" action="<? echo $PHP_SELF ?>">

Try that and see what happens

alternative

i make everything like that....only the query button came out.

<form method="POST" action="<? echo $PHP_SELF ?>"> 
<?
$conn = mysql_connect("localhost", "webauth", "webauth");   

$db = mysql_select_db("ddm");   

// running sql query  table = category, attributes = categoryID, cate 
$list = mysql_query("SELECT * FROM category", $conn); //change here    

while ($row = @mysql_fetch_assoc($list))  

{ //option value everything is here, query as much as data from database 
echo "<option value=\"$row[categoryID]\">$row[cate]</option>\n<br>";  

}  

if ($_POST['categoryID'] !=" ")   

{  

    $list2 = mysql_query("SELECT * FROM category2 where categoryID=$categoryID", $conn);//change 
    while ($row2 = @mysql_fetch_assoc($list2))   

    {  

    echo "<option value=\"$row2[categoryID]\">$row[cate2]</option>\n<br>";   

    }   

}
?>
</select>
<input type="submit" value="query" name="B1"> <!-- submit button--> 
</form>

-----------------------

kevinsequeira

u have categoriID above, i guess its a typo
if ure action page is abc.php?categoryID=1 then shouldnt this

$_POST['categoryID'] !=" "

$_GET['categoryID'] !=" "

reg
kevin

alternative

i still dun understand
do you have some example for this code?

kevinsequeira

ure select box has no name
<select>

change it to
<select name="categoryID"> if thats what it is.

also why do u have the post action=<?$_POST['categoryID']?>

im not sure what it is that you want to do, could u explain
are u submitting this form to another page, what is this page called ??

what do u mean by this ??
reg
kevin

alternative

actually i want to select the data from database
and then select for the second tim according to the 1st time data

if i use dropdownmenu to select 'general', then the second dropdownmenu will prompt out all related to 'general' element.

like that,

thanx in advance

alternative

anyone can help me on this??

bad76

Hi,

try this and report error...

<form method="POST" action="<? echo $PHP_SELF ?>"> 
<select>
<? 
$conn = mysql_connect("localhost", "webauth", "webauth");    

$db = mysql_select_db("ddm") or die("error 1:".mysql_error());    

$list = mysql_query("SELECT * FROM category", $conn)  or die("error 2:".mysql_error());    

while ($row = @mysql_fetch_assoc($list))   

{ 
echo "<option value=\"$row[categoryID]\">$row[cate]</option>\n<br>";   

}   

if ($_POST['categoryID'] !=" ")    

{   

    $list2 = mysql_query("SELECT * FROM category2 where categoryID=$categoryID", $conn) or die("error 3:".mysql_error());    

while ($row2 = @mysql_fetch_assoc($list2))    
{   
echo "<option value=\"$row2[categoryID]\">$row[cate2]</option>\n<br>";    
}    
} 
?> 
</select> 
<input type="submit" value="query" name="B1"> <!-- submit button--> 
</form>

I hope this help you...

alternative

there is nothing out...just a empty dropdowmenu prompt out

<form method="POST" action="/dropdownmenu/ddm5.php">
<select>
error 1:Access denied for user: 'webauth@localhost' to database 'ddm'

bad76

error 1:Access denied for user: 'webauth@localhost' to database 'ddm'

and this is nothing ?!?

This is the problem: the user/password you use don't have permission to select the requested db. You can't access it.

Because is on localhost may be you have used wrong user...

alternative

then how i gonna know how to check the username and password?

bad76

By default the user is 'root' and pwd is empty ''

else go in phpMyAdmin and look at user table...

alternative

i try to change it to root
and then leave the username and password become blank..but still not working.

bad76

Try with phpMyadmin,
there is the user list. Look for a user and password in the list...

alternative

i am using phpmyadmin..but i dunno how to check the username.

bad76

There is some
different version of phpmyadmin, so i can't help much.

You have to find in db 'mysql' the 'user' table. Then click on show and in field user & password you can read the values you need.

alternative

i change to code username and password to correct one.

<form name="test_form" action="<? echo $PHP_SELF; ?>" method="POST"> 
<?php 
//db connections
$conn = mysql_connect("localhost","root","");     

$db = mysql_select_db("ddm") or die("error 1:".mysql_error());    

if(!isSet($_POST['ddm1'])){ 
        $ddm1 = $_POST['ddm1']; 
        $result = mysql_query("select * from category"); 
        echo "<select name=ddm1 onChange=\"java script:document.test_form.submit();\">"; 
  while ($row = mysql_fetch_array($result)){ 
    echo "<option value=".$row['categoryID']; 
        echo ">".$row['cate']."</option>"; 
  } 
        echo "</select>";     

}else{ 
        $result = mysql_query("select * from category"); 
        echo "<select name=ddm1 onChange=java script:\"document.test_form.submit();\">"; 
  while ($row = mysql_fetch_array($result)){ 
    echo "<option value=".$row['categoryID']; 
            if($ddm1 == $row['cate']) 
                echo " SELECTED"; 
         echo ">".$row['cate']."</option>"; 
  } 
        echo "</select>";
        $result = mysql_query("Select * from category2 where categoryID= $ddm1"); 
        echo "<select name=ddm2>"; 
  while ($row = mysql_fetch_array($result)){ 
   $option .= "<option value=".$row['category2ID'].">".$row['cate2']."</option>"; 
  } 
        echo "</select>";
		}         

?> 
</form>

but bad76, the things din works. just a dropdownmenu prompt out..and nothing happened after i select something.